Concept explainers
(a)
To find: the value of the mean adhesion of the locomotive
(a)
Answer to Problem 57E
The mean adhesion should be 0.382
Explanation of Solution
Given:
The standard deviation remains at
Calculation:
Solve for the appropriate mean of the distribution so that the adhesion is less than 0.30 on less than
Find the z -value for which
Using Standard normal table, the z -value corresponding to 0.02,
Since this z -value should correspond to an adhesion of 0.30,
Hence, the mean adhesion should be 0.382
Conclusion:
Hence, the mean adhesion should be 0.382
(b)
To find: the value of the standard deviation of the adhesion
(b)
Answer to Problem 57E
The standard deviation of the adhesion values should be 0.034.
Explanation of Solution
Given:
The mean adhesion stays at
Calculation:
If the mean adhesion stays at 0.37, find the standard deviation that is to be decreased to ensure that the train will arrive late less than
To find the standard deviation,
The standard deviation of the adhesion values should be 0.034.
Conclusion:
Therefore, the standard deviation of the adhesion values should be 0.034.
(c)
To find: whether option a or b is preferable.
(c)
Answer to Problem 57E
Part (b) is preferred.
Explanation of Solution
Given:
The mean adhesion stays at
Calculation:
Compare the two options in (a) and (b) to find which option is preferable. Our objective is to decrease the variability in adhesion from trip to trip.
To compare the options, find the area under the
Using standard normal table, find that the area to the right of 0.5 as follows:
That is, using
Taking the standard deviations as 0.034, calculate the z value as follows:
Using standard normal table, find that the area to the right of 0.5 as follows:
That is, using
Since very less percentage of proportions isto the right of 0.5 when
Conclusion:
Therefore, part (b) is preferred.
Chapter 2 Solutions
The Practice of Statistics for AP - 4th Edition
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